MHB S3 Group Symmetry: $(xy)^2 \ne x^2y^2$ Example

[Guest2]

Problem: In $S_3$ give an example of two elements $x$ and $y$ such that $(xy)^2 \ne x^2y^2$.

Attempt: Consider the mapping $\phi: x_1 \mapsto x_2, x_2 \mapsto x_1, x_3 \mapsto x_3$ and the mapping $\psi: x_1 \mapsto x_2, x_2 \mapsto x_3, x_3 \mapsto x_1$. We have that the elements $\phi, \psi \in S_3$. We also have $x_1 (\phi \psi)^2 = x_1\phi \psi \phi \psi = x_3 \phi \psi = x_1.$ We also have that $\phi^2 = e$ and thus $ x_1 \phi^2 \psi^2 = x_1 \psi^2 = x_2 \psi = x_3$. Therefore $(\phi \psi)^2 \ne \phi^2 \psi^2$. Hence it's enough to take $x = \phi$ and $y = \psi$.

Is the above any good?

 

[Deveno]

Problem: In $S_3$ give an example of two elements $x$ and $y$ such that $(xy)^2 \ne x^2y^2$.

Attempt: Consider the mapping $\phi: x_1 \mapsto x_2, x_2 \mapsto x_1, x_3 \mapsto x_3$ and the mapping $\psi: x_1 \mapsto x_2, x_2 \mapsto x_3, x_3 \mapsto x_1$. We have that the elements $\phi, \psi \in S_3$. We also have $x_1 (\phi \psi)^2 = x_1\phi \psi \phi \psi = x_3 \phi \psi = x_1.$ We also have that $\phi^2 = e$ and thus $ x_1 \phi^2 \psi^2 = x_1 \psi^2 = x_2 \psi = x_3$. Therefore $(\phi \psi)^2 \ne \phi^2 \psi^2$. Hence it's enough to take $x = \phi$ and $y = \psi$.

Is the above any good?

That looks good, although you should make it explicit that $\phi\psi$ is the mapping:

$x_1 \mapsto x_3$
$x_2 \mapsto x_2$
$x_3 \mapsto x_1$.

When you learn cycle notation, these kinds of problems will be much easier to write about.

(and yes, to show that two mappings $f,g:A \to B$ are unequal, it is enough to exhibit a single $a \in A$ such that $(a)f \neq (a)g$...in this case we have $A = B = \{x_1,x_2,x_3\}$).